In a 2.0 molal aqueus solution of a weak acid HX the degree of disssociation is 0.25. The freezing point of the solution will be nearest to: (`K_(f)=1.86 K kg "mol"^(-1)`)

Options

A. `-0.26^(@)C`
B. `0.465^(@)C`
C. `-0.48^(@)C`
D. `4.65^(@)C`

Answer

To solve the problem, we need to calculate the freezing point depression of a 2.0 molal aqueous solution of a weak acid HX, given that the degree of dissociation (α) is 0.25 and the freezing point depression constant (Kf) is 1.86 K kg/mol.

### Step 1: Determine the Van't Hoff factor (i)
The weak acid HX dissociates into H⁺ and X⁻ ions. The dissociation can be represented as:
\[ HX \rightleftharpoons H^+ + X^- \]

Initially, we have 1 mole of HX. At equilibrium, if α is the degree of dissociation:
- Moles of HX remaining = \( 1 - \alpha \)
- Moles of H⁺ formed = \( \alpha \)
- Moles of X⁻ formed = \( \alpha \)

Thus, the total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

Substituting α = 0.25:
\[ \text{Total moles} = 1 + 0.25 = 1.25 \]

The Van't Hoff factor (i) is given by:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{1 + \alpha}{1} = 1 + 0.25 = 1.25 \]

### Step 2: Calculate the depression in freezing point (ΔTf)
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where:
- \( K_f = 1.86 \, \text{K kg/mol} \)
- \( m = 2.0 \, \text{molal} \)
- \( i = 1.25 \)

Substituting the values:
\[ \Delta T_f = 1.86 \cdot 2.0 \cdot 1.25 \]

Calculating:
\[ \Delta T_f = 1.86 \cdot 2.0 = 3.72 \]
\[ \Delta T_f = 3.72 \cdot 1.25 = 4.65 \, \text{K} \]

### Step 3: Determine the freezing point of the solution
The freezing point of pure water (T₀f) is 0 °C. The freezing point of the solution (Tf) can be calculated as:
\[ T_f = T_{0f} - \Delta T_f \]
\[ T_f = 0 - 4.65 = -4.65 \, \text{°C} \]

### Final Answer
The freezing point of the solution will be nearest to **-4.65 °C**.

---